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3 years ago

Which type of tissue allows for cordination and control movement ?

Physics

2 answers:

AnnZ [28]3 years ago

5 0

Nervous tissue is used for coordination and control movement.

<u>Explanation: </u>

The group of organized cells that is in the nervous system, which is responsible in controlling the movements of the human body, sending and carrying signals to and from different body parts is called Nervous tissue. It is also responsible in controlling functions in the body like digestion.

This tissue is classified into categories such as neurons and neroglia. The electric impulses are transmitted by neurons and supporting and protecting neurons is done by neuroglia. The whole nervous system is comprised of neurons.

Rom4ik [11]3 years ago

4 0

The Nerve Tissue allows for coordination and control movement

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a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat

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1) The minimum force needed is 920 N

2) The minimum force is 460 N

3) The minimum force is 598 N

Explanation:

We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:

Its weight, W, acting downward
The force applied to lift it, F, upward

So the net force on the railroad is

$\sum F = F - W$

where

W = 920 N is the weight of the railroad

In order to lift the railroad, the net force must be upward, so

$\sum F \geq 0$

And therefore

$F\geq W$

which means that the minimum force needed is equal to the weight of the railroad, 920 N.

In this case, we have to use the principle of equilibrium of moments.

In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:

$W \frac{L}{2} = F L$

where

W = 920 N is the weight

L = 2.6 m is the length of the railroad

F is the force applied

We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.

Solving for F,

$F=\frac{W}{2}=\frac{920}{2}=460 N$

This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.

This means that the distance between the point of application of the force F and the pivot is

L' = L - 0.6

where

L = 2.6 m

Therefore the equation for the equilibrium of moments becomes

$W\frac{L}{2}=F(L-0.6)$